Class XII · Chemistry
Principles Related to Practical Chemistry: NEET MCQ practice
138 practice questions on Principles Related to Practical Chemistry, each with an explanation. Practise them chapter by chapter in the GyaanLoop app.
138
Practice questions
84
Easy
50
Medium
4
Hard
What Principles Related to Practical Chemistry covers
Topics in the NEET (UG) 2026–27 syllabus for this chapter.
- Detection of extra elements (nitrogen, sulphur, halogens) in organic compounds
- Detection of functional groups- hydroxyl, carbonyl, carboxyl and amino groups
- Preparation of inorganic and organic compounds (Mohr's salt, potash alum, acetanilide, iodoform, etc.)
- Titrimetric exercises- acids, bases, indicators; oxalic acid vs KMnO4; Mohr's salt vs KMnO4
- Chemical principles involved in qualitative salt analysis (cations and anions)
- Enthalpy of solution of CuSO4, enthalpy of neutralization, lyophilic and lyophobic sols
- Kinetic study of reaction of iodide ions with hydrogen peroxide at room temperature
Sample questions on Principles Related to Practical Chemistry
Try each one, then open the answer and explanation.
Question 1 · Practice question
Roasting is done generally in case of the
- A.Silicate ores
- B.Sulphide ores
- C.Oxide ores
- D.Carbonate ores
Show answer
Answer: B. Sulphide ores
Roasting is a process specifically used for sulphide ores to convert them into oxides by heating in air. For carbonate ores, calcination is used, and oxide ores do not require such conversion. This is a straightforward recall from NCERT metallurgy concepts.
Question 2 · Practice question
One Integer Value Correct Type Direction (Q. Nos. 21-24) This section contains 4 questions. When worked out will result in an integer from 0 to 9 (both inclusive). Q. Number of alloys that contain nickel among the following solder, gun metal, German silver, nichrome, monel metal, constanton, bell metal, duralumin, type metal, invar, alnico.
- A.6
- B.7
- C.5
- D.3
Show answer
Answer: A. 6
You need to recall the composition of each listed alloy. Nickel-containing alloys are German silver (Cu-Zn-Ni), nichrome (Ni-Cr-Fe), monel metal (Ni-Cu), constantan (Cu-Ni), invar (Fe-Ni), and alnico (Al-Ni-Co-Fe), totaling 6. A common mistake could be missing constantan or including duralumin (Al-Cu-Mg-Mn, no nickel), which would change the count.
Question 3 · Practice question
Green vitriol is FeSO₄ .xH 2 O and white vitriol is ZnSO₄ .yH 2 O, Then, the values of x and y are
- A.8
- B.5
- C.4
- D.7
Show answer
Answer: D. 7
Green vitriol is the common name for ferrous sulfate heptahydrate (FeSO₄·7H₂O), and white vitriol is zinc sulfate heptahydrate (ZnSO₄·7H₂O), so both x and y are 7. This is a straightforward recall question from inorganic salt nomenclature. The only slight nuance is that students might confuse it with blue vitriol (CuSO₄·5H₂O) or Epsom salt (MgSO₄·7H₂O), but the names 'green' and 'white' vitriol are specifically associated with the heptahydrates of iron(II) sulfate and zinc sulfate.
Question 4 · Practice question
A mixed solution of potassium hydroxide and sodium carbonate required 15 mL. of an N/20 HCI solution when titrated with phenolphthalein as an indicator. But the same amount of the solution when titrated with methyl orange as an indicator required 25 mL of the same acid. The amount of KOH present in the solution is
- A.0.014g
- B.0.14g
- C.0.028g
- D.1,4 g
Show answer
Answer: A. 0.014g
In a mixture of KOH (strong base) and Na2CO3 (weak base), phenolphthalein indicates neutralization of KOH plus conversion of CO3²- to HCO3-, while methyl orange indicates complete neutralization to H2CO3. By setting up equations for acid used at each end point, moles of KOH come out as 0.00025 mol, giving mass = 0.00025 × 56 = 0.014 g. Option (c) 0.028 g would be the result if carbonate were incorrectly treated as using half the acid in methyl orange titration.
Question 5 · Practice question
Mohr’s salt is a double salt with the formula FeSO₄ (NH₄)₂SO₄ xH 2 O. Here, value of x is
- A.9
- B.7
- C.5
- D.6
Show answer
Answer: D. 6
Mohr's salt is ferrous ammonium sulfate hexahydrate, FeSO₄·(NH₄)₂SO₄·6H₂O, so x = 6. Students might confuse it with ferrous sulfate heptahydrate (FeSO₄·7H₂O), but the double salt has a fixed composition of six water molecules.
More Class XII Chemistry chapters
- Solutions407 questions
- Electrochemistry309 questions
- Chemical Kinetics223 questions
- P-Block Elements389 questions
- d- and f-Block Elements293 questions
- Co-ordination Compounds310 questions